JEE MAIN - Physics (2023 - 12th April Morning Shift - No. 5)
The ratio of escape velocity of a planet to the escape velocity of earth will be:-
Given: Mass of the planet is 16 times mass of earth and radius of the planet is 4 times the radius of earth.
$$1: 4$$
$$1: \sqrt{2}$$
$$4: 1$$
$$2: 1$$
Explanation
The escape velocity of a planet or a celestial body is given by:
$$v_e = \sqrt{\frac{2GM}{r}}$$
where $$G$$ is the gravitational constant, $$M$$ is the mass of the planet, and $$r$$ is the radius of the planet.
Let the subscripts "p" and "e" denote the planet and earth, respectively. Then, we have:
$$\frac{v_{e,p}}{v_{e,e}} = \frac{\sqrt{\frac{2G M_p}{r_p}}}{\sqrt{\frac{2G M_e}{r_e}}} = \sqrt{\frac{M_p r_e}{M_e r_p}}$$
Substituting the given values, we get:
$$\frac{v_{e,p}}{v_{e,e}} = \sqrt{\frac{16 \cdot 1}{1 \cdot 4}} = \sqrt{4} = 2$$
Therefore, the ratio of the escape velocity of the planet to that of earth is 2 : 1.
$$v_e = \sqrt{\frac{2GM}{r}}$$
where $$G$$ is the gravitational constant, $$M$$ is the mass of the planet, and $$r$$ is the radius of the planet.
Let the subscripts "p" and "e" denote the planet and earth, respectively. Then, we have:
$$\frac{v_{e,p}}{v_{e,e}} = \frac{\sqrt{\frac{2G M_p}{r_p}}}{\sqrt{\frac{2G M_e}{r_e}}} = \sqrt{\frac{M_p r_e}{M_e r_p}}$$
Substituting the given values, we get:
$$\frac{v_{e,p}}{v_{e,e}} = \sqrt{\frac{16 \cdot 1}{1 \cdot 4}} = \sqrt{4} = 2$$
Therefore, the ratio of the escape velocity of the planet to that of earth is 2 : 1.
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